One Way ANOVA Table for N = 18, k = 3 & α = 5%
One way ANOVA table or summary to estimate the test of hypothesis (H0) at 5% or 0.05 significance level for sample means by using F-distribution table
| One-Way ANOVA Classification Table | ||||
|---|---|---|---|---|
| Sources of Variation | df | SS | MSS | F-Ratio |
| between Treatments | 2 | 10.4111 | 10.4111/2 = 5.2056 | 5.2056/3.2467 = 1.6034 |
| Error | 15 | 48.7 | 48.7/15 = 3.2467 | |
| Total | 17 | |||
Work with Steps for N = 18, k = 3 & α = 5%
The below is the example work with steps shows how to generate the one way ANOVA table for the different means of treatments to find the F test statistic to test hypothesis (H0) at a 5% or 0.05 significance level for three sample means x1, x2 & x3.
Workout :
step 1 Address the formula input parameters and values
x1 = 10, 12, 13, 11, 10, 13 & 15
x2 = 9, 11, 13, 9 & 11
x3 = 11, 15, 13, 12, 14 & 10
step 2 Form the below table to carry out the analysis of variance
Individual Squares :
step 3 Find Correction Factor (CF)
N = total number of elements in all three sample dataset.
N = 7 + 5 + 6
N = 18
Correction factor (CF) = G2N
CF = 212218
= 44944/18
CF = 2496.8889
step 4 Find the sum of squares of all individual items (∑xij) and then total sum of squares (TSS)
Total sum of squares (TSS) = ∑∑xij2 - CF
= 2556 - 2496.8889
TSS = 59.1111
step 5 Find the sum of squares between the treatments (SST)
SST = k ∑ i = 1 Ti2ni - CF
= (∑x1)2n1+(∑x2)2n2+(∑x3)2n3 - CF
= 7056/7+2809/5+5625/6 - 2496.8889
= (1008 + 561.8 + 937.5) - 2496.8889
= 2507.3 - 2496.8889
= 10.4111
step 6 Find sum of squares due to the error (SSE)
SSE = TSS - SST
= 59.1111 - 10.4111
= 48.7
step 7 The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table.
One-Way ANOVA Classification Table
step 8 Find Critical value of Fe or Table value of F
The critical value of Fe from the F-distribution table for the degrees of freedom (k - 1, N - k) at 5% significance level.
For treatments, the critical value Fe(2,15) from F-distribution table at 5% of significance level is 3.6823
Inference
There is no significance difference between treatments, since the calculated value of F0 = 1.6034 is smaller than the table value of Fe = 3.6823. Therefore the null hypothesis H0 is accepted.
step 1 Address the formula input parameters and values
x1 = 10, 12, 13, 11, 10, 13 & 15
x2 = 9, 11, 13, 9 & 11
x3 = 11, 15, 13, 12, 14 & 10
step 2 Form the below table to carry out the analysis of variance
| Total | square | ||||||||
| x1 | 10 | 12 | 13 | 11 | 10 | 13 | 15 | ∑x1 = 84 | (∑x1)2 = 7056 |
| x2 | 9 | 11 | 13 | 9 | 11 | ∑x2 = 53 | (∑x2)2 = 2809 | ||
| x3 | 11 | 15 | 13 | 12 | 14 | 10 | ∑x3 = 75 | (∑x3)2 = 5625 | |
| G = 212 | ∑Ti2 = 15490 | ||||||||
Individual Squares :
| Total | ||||||||
| x12 | 100 | 144 | 169 | 121 | 100 | 169 | 225 | ∑x12 = 1028 |
| x22 | 81 | 121 | 169 | 81 | 121 | ∑x22 = 573 | ||
| x32 | 121 | 225 | 169 | 144 | 196 | 100 | ∑x32 = 955 | |
| ∑xij2 = 2556 | ||||||||
N = total number of elements in all three sample dataset.
N = 7 + 5 + 6
N = 18
Correction factor (CF) = G2N
CF = 212218
= 44944/18
CF = 2496.8889
step 4 Find the sum of squares of all individual items (∑xij) and then total sum of squares (TSS)
Total sum of squares (TSS) = ∑∑xij2 - CF
= 2556 - 2496.8889
TSS = 59.1111
step 5 Find the sum of squares between the treatments (SST)
SST = k ∑ i = 1 Ti2ni - CF
= (∑x1)2n1+(∑x2)2n2+(∑x3)2n3 - CF
= 7056/7+2809/5+5625/6 - 2496.8889
= (1008 + 561.8 + 937.5) - 2496.8889
= 2507.3 - 2496.8889
= 10.4111
step 6 Find sum of squares due to the error (SSE)
SSE = TSS - SST
= 59.1111 - 10.4111
= 48.7
step 7 The above sum of squares together with their respective degrees of freedom and mean sum of squares will be summarized in the following table.
| Sources of Variation | df | SS | MSS | F-ratio |
| between Treatments | k - 1 | SST | SST/k-1 = MST | MST/MSE = F T |
| Error | N - k | SSE | SSE/N-k = MSE | |
| Total | N - 1 |
One-Way ANOVA Classification Table
| Sources of Variation | df | SS | MSS | F-Ratio |
|---|---|---|---|---|
| between Treatment | 3-1 = 2 | 10.4111 | 10.4111/2 = 5.2056 | 5.2056/3.2467 = 1.6034 |
| Error | 18-3 = 15 | 48.7 | 48.7/15 = 3.2467 | |
| Total | 18-1 = 17 |
step 8 Find Critical value of Fe or Table value of F
The critical value of Fe from the F-distribution table for the degrees of freedom (k - 1, N - k) at 5% significance level.
For treatments, the critical value Fe(2,15) from F-distribution table at 5% of significance level is 3.6823
Inference
There is no significance difference between treatments, since the calculated value of F0 = 1.6034 is smaller than the table value of Fe = 3.6823. Therefore the null hypothesis H0 is accepted.
